How do you pick resistors to adjust bias voltage?

[momo]

In the spirit of trying to come up with some "new" effect,or effect of my own, Id like to experiment with simple layout say 2 trannys and a battery to learn how to choose which resistor values are needed to achieve the desired bias voltages.
Sorry if this is a obvious question, I did a search and found alot of good info, but none on this question.
I am aware that one member of the forum put together a software to calculate those for Bigg muff and FFace. But id like to know the logic behind the calculations.
thanks!

[R.G.]

The short version:
1. Use a high gain bipolar transistor. That lets you ignore base current until you want to mess with it. Very handy. Hfe >100 is good.
2. Start at the collector and emitter. Of your power supply voltage (e.g. 9V) apportion the amount of power supply voltage you need to the collector and base for the application. For instance, in an emitter follower, you want the emitter nearly at the middle of the power supply (actually, 1/2 power supply plus 1/2 of the base-emitter drop), so the emitter would be at 4.5V.

For an amplifier, you want the maximum possible voltage split between the collector resistor and the collector-emitter. The voltage on the emitter should be large compared to the variation in the base-emitter voltage; making it 1x Vbe works fine. Subtract the emitter voltage from the total power supply voltage, then divide by two. The result is the voltage across the collector resistor.

Choose an operating current. Something between 10uA and 1ma is normal for effects gain stages. 100uA is a good place to start. Then the collector resistor is the voltage you previously calculated for the collector resistor divided by this current. The emitter resistor is the voltage you chose for the emitter divided by this current.

3. Set the base voltage. You know the emitter voltage. The base must be one Vbe higher. Vbe varies. It can be as little as 0.45V for uA-collector currents up to 0.7V for high currents. 0.55 is a good place to start. Add 0.55V to the emitter voltage and that's the base voltage.
4. Calculate the biasing resistors. You know what voltage the base must sit at from 3. You know the power supply voltage. Design the two resistors. The output voltage is Vbase = 9V*R2/(R1+R2) where R1 is the top resistor and R2 is the bottom resistor to ground. That's not enough information to solve for R1 and R2. So you have to pick the current through them. You just choose the current you will let flow through the two, by I = 9V/(R1+R2).  Make this current greater than 10x the base current, preferably 50-100 times the base current. The base current is the collector current you picked divided by hfe. It will be small. You do not need to know EXACTLY the hfe, only approximately, so guess at a nominal hfe. It won't matter much, as this just chooses about what resistance is needed.

Once you have selected a current for the input dividers R1 and R2, you know R1+R2 = 9V/I and  you can substitute this number into the previous equation: R2= Vbase*(9V/I)/9V = Vbase/I . Then R1 = (9V-Vbase)/I


Raising Vb raises the emitter voltage; this increases the current through the emitter resistor; this same current flows through the collector resistor and lowers the collector. Lowering Vb does the reverse.

[momo]

R.G....Im always amazed at the level of info and time you take for the responses!, either you type amazingly fast, or you got the patience of a Monk!....

Thanks again for this info, Ill take the next hour and digest that....oh and since your reading, I recently got a hold of your article on the Rangemaster...I built this one and tried different Ge's till I found one I liked.That was before learning how to do your test on trannys.
Anyway what im saying is that I realize that my Rangemaster has lots more in potential so I want to use it to learn how to bias.

In the past 2 months that I discovered this forum, my tone has gone from what I thought was good(it was pretty lame and clinical), to heavy thick analog funk!.....This has in turn augmented my joy of playing,which makes me play more, which makes me a better guitarist.....the advantages are endless!
You guys have really put my craft on another level and I will forever be greatfull.
Now if I can only start on my own stuff, I will give it back to you guys hoping you like the new edition!

[momo]

OK, basic questions for now would be:

Am I right in saying that in a emitter follower mode, the tranny does not nessesarily amplify to the next stage but rather would add/change/modify the signal in a way, without amplifying to the next stage? as in the Neovibe phase stages?

If that is right then I guess a amplifier stage would be like in the Bigg Muff,say maybe the last 3 trannys, where the gain gets higher at every stage?

I guess I understand that raising the Vb would raise the emmitor(in an amplifier mode). But why would we want to raise the collector voltage?, is that how you make it a emitter follower?

Thaaannnks alot....

[rockgardenlove]

"2. Start at the collector and emitter. Of your power supply voltage (e.g. 9V) apportion the amount of power supply voltage you need to the collector and base for the application. For instance, in an emitter follower, you want the emitter nearly at the middle of the power supply (actually, 1/2 power supply plus 1/2 of the base-emitter drop), so the emitter would be at 4.5V."
How do you know how much you need for the collector+base?  Would you determine this by what sort of gain you want the gainstage you half?  Because the larger the voltage drop between B+ and the emitter the larger the voltage swing, right?



I went ahead with my (probably faulty) logic, and tried to make a max-gain gainstage.  I therefore took the emitter straight to ground.  (9v-0)/2 is of course 4.5, so my collector voltage would be 4.5v.  I used your 100uA current, so the collector resistor would be .045Ω, which is *really* impractical.  I must have messed up.  Still, I pressed onward:  The base voltage would have to be .55v (using your default again).  Through some calculating I figured that R1 would come out to .56Ω and R2 would come out to be .037Ω.  Now, somewhere in there I got messed up.  So I headed over here to test it:
http://ourworld.compuserve.com/homepages/Bill_Bowden/r2.htm
After plugging in my resistor values, and leaving the current blank, I came up with 15A instead of .000015A like I should have...
This makes it seem to me like I just got really mixed up with the units somewhere. 

So, to summarize:
R1=.563Ω
R2=.037Ω
Collector resistor=.045Ω
Emitter resistor=0Ω
Battery voltage=9v

Is there any hope at all this would bias up?
I think I got my units screwy in there somewhere.  Would they be correct if you shifted the decimal point over to the right a bit?

Thanks very much!

[rockgardenlove]

Bump?

Any help appreciated!

[R.G.]

Am I right in saying that in a emitter follower mode, the tranny does not nessesarily amplify to the next stage but rather would add/change/modify the signal in a way, without amplifying to the next stage? as in the Neovibe phase stages?

Yes. The emitter MUST follow (!!! emitter follower !!!) the base. So the gain is very near 1.000. It's actually 0.99 or so for a good NPN bipolar. The voltage is the same, but the emitter follower can source much more current, so it buffers the signal.

If that is right then I guess a amplifier stage would be like in the Bigg Muff,say maybe the last 3 trannys, where the gain gets higher at every stage?

Actually, all four transistors in the BMP amplify, but yes, that's correct.

I guess I understand that raising the Vb would raise the emmitor(in an amplifier mode). But why would we want to raise the collector voltage?, is that how you make it a emitter follower?

In an emitter follower, you want to make the maximum possible signal come out of the emitter. If there is a resistor in the collector, then you can saturate the transistor by pulling so much current that the voltage between collector and emitter goes to nearly zero, and then the transistor clips. By moving the collector voltage up (and making the collector resistor small or 0) you give the emitter more voltage to wiggle up and down, and therefore can follow bigger voltages without clipping.

How do you know how much you need for the collector+base?  Would you determine this by what sort of gain you want the gainstage you half?  Because the larger the voltage drop between B+ and the emitter the larger the voltage swing, right?

The voltage between collector and base is kind of uninteresting. What matters is that Vbe is always about one diode drop, and that the emitter follows the base. All the rest of the circuit comes out of those issues. The voltage between collector and base varies as the collector wiggles around. Actually, the collector does not wiggle around. The emitter follows the base, letting the signal current through, and that same current flows in the collector resistor. The collector resistor voltage wiggles around.

I went ahead with my (probably faulty) logic, and tried to make a max-gain gainstage.  I therefore took the emitter straight to ground.  (9v-0)/2 is of course 4.5, so my collector voltage would be 4.5v. 

So far so good, but you've already built in a problem for yourself. With the emitter straight to ground, the emitter can't follow the base. ACK!

What actually happens is that there is an internal junction resistor that acts like it's on the emitter, but inside the transistor where you can't get at it. This is the "Shockley resistance" and is about 25mV/Ie. I didn't cover that part in my polemic, but it's critical when you ground the emitter.

I used your 100uA current, so the collector resistor would be .045Ω, which is *really* impractical.  I must have messed up.

Yep, you dropped a decimal. 4.5V/0.0001 = 45,000, not .0452. The collector resistor is 4.5V at a current of 100uA.

Still, I pressed onward:  The base voltage would have to be .55v (using your default again).  Through some calculating I figured that R1 would come out to .56Ω and R2 would come out to be .037Ω.  Now, somewhere in there I got messed up.  So I headed over here to test it:
http://ourworld.compuserve.com/homepages/Bill_Bowden/r2.htm
After plugging in my resistor values, and leaving the current blank, I came up with 15A instead of .000015A like I should have...
This makes it seem to me like I just got really mixed up with the units somewhere.

So, to summarize:
R1=.563Ω
R2=.037Ω
Collector resistor=.045Ω
Emitter resistor=0Ω
Battery voltage=9v

Is there any hope at all this would bias up?
I think I got my units screwy in there somewhere.  Would they be correct if you shifted the decimal point over to the right a bit?

Sorry, no hope at all. The emitter-at-ground is a special case. Sorry - I only gave the general case, not the special one.

With the emitter at ground, the base will always be at 0.45V to 0.7V. The signal voltage can only wiggle around about 25mV without distortion, right? Otherwise it would turn the base-emitter diode fully on and fully off. Another way to look at it is that with the emitter grounded, changing the base voltage by 25mV to 50mV drives the transistor fully on to fully off, so the gain is indeterminate, but is necessarily larger than 9V(the collector swing) divided by 25-50mv (the base swing) which is big.

What's really going to happen is that the internal Shockley resistance will determine the gain, but the base bias voltage will be determined by the base diode voltage, not what resistor divider you hang on it.

The grounded-emitter case is a very bad case for biasing. I'll post some more on that one when the ends of my fingers grow back.

[rockgardenlove]

Well, it looks like I followed the process somewhat correctly. 
I don't feel totally bewildered–always a good thing in my book.

Still about confused about this:
In your example, you said that 4.5v was a good voltage for an emitter follower on the emitter.  Why?  Say I want to make a standard old collector follower, what's a good emitter voltage to choose?  Anything over Vbe?  Bit confused here.

Thanks VERY MUCH RG!

[R.G.]

In your example, you said that 4.5v was a good voltage for an emitter follower on the emitter.  Why? 

In an emitter follower with the proposed 9V power supply, 4.5V is half the power supply. Half the power supply is a quick way to get the largest undistorted output signal. If the emitter is at 4.5V with no signal, the base at 5.1V, and the collector at 9V, the emitter can follow the base down until the emitter voltage goes to 0V, at which point the transistor is cut off, or up to just under 9V, at which point the transistor is saturated. The base moves along about 0.5V higher than the emitter.  So this setup can produce almost 9V peak to peak output signal without distorting. If the emitter is closer to either ground or the power supply than the middle, it bumps into the one it's closest to first, so the signal distorts earlier.

Say I want to make a standard old collector follower, what's a good emitter voltage to choose?  Anything over Vbe?  Bit confused here.

You should be. There is no "collector follower". The collector always moves the opposite polarity from the base.

[rockgardenlove]

Oops, I was under the impression that the output was just called the follower.  So, if I want the out of phase output form the collector, what do I want for the emitter voltage wise?

[mac]

I am aware that one member of the forum put together a software to calculate those for Bigg muff and FFace. But id like to know the logic behind the calculations.
thanks!

It's embarrasing for me to admit that I can tell you about the physics behind a BJT, Fet or valve, or solve the dc bias of the FF, BMP and voltage divider and write a piece of code... but the truth is that a couple of years ago I did not have any idea of how to put a transistor to work!!!
Now in my experience, this forum is more instructive than zillions of pages I read about biasing transistors, dc and ac analysis, etc. Mainly because it is interactive and there are a lot of kind members that answer your question regardless if one is a newbie or not. And most important, the answers I got or read here are direct, clear and simple. I mean, two books may talk about the same subject but one is easy to understand than the other.
Because of the positive feedback I wrote that code. Now momo has given me the idea of adding a help window with some pratical examples on how to select resistors, working voltages, clipping, etc., maybe RG words ... if you agree of course.

mac

[rockgardenlove]

Oops, I was under the impression that the output was just called the follower.  So, if I want the out of phase output form the collector, what do I want for the emitter voltage wise?

Bump

[R.G.]

Oops, I was under the impression that the output was just called the follower. 

Ah, OK. No biggie. Nomenclature problem.

So, if I want the out of phase output form the collector, what do I want for the emitter voltage wise?

You want as much of the available voltage as is "reasonable" (to be defiined later) to be divided equally between the collector resistor and the transistor collector-to-emitter. The phrase divided equally means that whatever the emitter doesn't use, the collector sits at half that. If the power supply is 9V, the emitter is set to 1V, then the collector voltage should be about 5V. I say "about" because the emitter could actually go all the way to ground, but it's best to plan on only taking off +/-4V, not 4.5V.

The DC emitter voltage only really needs to be big compared to the change in Vbe due to temperature. In general, an emitter voltage of 0.5V to 1V is conservative. 0.1V to 0.5V is less conservative.

The bigger the emitter voltage is compared to the base-emitter diode drop, the more stable the DC bias is against changes in temperature and hfe. But you pay for the stability with lowered DC gains and smaller available signal swing.

[rockgardenlove]

Hmmm...so if you want the emitter to be 1v, that leaves 8v to be split between the collector and the transistor's channel?  So wouldn't you want the collector to be 4v?

[R.G.]

Hmmm...so if you want the emitter to be 1v, that leaves 8v to be split between the collector and the transistor's channel?  So wouldn't you want the collector to be 4v?

Remember, we're counting volts up from ground.

Emitter is 1V up. Four volts up from there is the collector, total of 5V. Then four volts up from there is 9V, the power supply. The collector splits the available 8V out of the total 9V.

[rockgardenlove]

Right!  In that case it makes perfect sense!  I was actually thinking that in my head!  But I didn't say it cause I didn't want to look like a fool.

[rockgardenlove]

Alright, how about this then:
R1-496.7k
R2-103.3k
Collector resistor-40k
Emitter resistor-10k

Emitter voltage-1V
Collector voltage-4V
Base voltage-1.55V
HFE-500

Would that bias up?



Simulation:

Wave on the left is the output...doesn't look right to me?

[rockgardenlove]

Bump!

[momo]

Well, Im certainly not at RockGardens level and lost the pace early on....no prob, this will be my tutorial that I will follow at my pace.
Thanks R.G for the effort and info!

[rockgardenlove]

Sorry...I totally hijacked your thread.  Should I start a new one?

Alright, how about this then:
R1-496.7k
R2-103.3k
Collector resistor-40k
Emitter resistor-10k

Emitter voltage-1V
Collector voltage-4V
Base voltage-1.55V
HFE-500

Would that bias up?



Simulation:

Wave on the left is the output...doesn't look right to me?

Help still appreciated! 
Thanks RG!